Question 2 of 6: Function realization — gates, MUX, decoder

Question 2: Function realization — gates, MUX, decoder (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three-variable functions specified by minterm/maxterm lists and by algebraic form; for $f_1$ the combinations $ABC=011$ (m3) and $111$ (m7) are don’t-cares.

Find. 8:1-MUX realizations for part (a), 4:1-MUX realizations for part (b), and a shared 3:8-decoder + OR realization for part (c).

Approach. An 8:1 MUX with $ABC$ on its select lines reproduces the truth table directly (data input $I_k$ = value of minterm $k$, don’t-cares free to tie either way); for the 4:1 MUX use two variables as the select lines and express the residue on the third variable as each data input; for the decoder, a 3:8 decoder generates every minterm, so each output function is simply the OR of its minterms.

1. Part (a), 8:1 MUX. Connect $S_2S_1S_0=A\,B\,C$ on each MUX and tie data input $I_k$ to the function value on row $k$. For $f_1=\Sigma m(1,4,6)$ with don’t-cares m3, m7: $I_1=I_4=I_6=1$, $I_0=I_2=I_5=0$, and $I_3,I_7$ may be tied to either rail (tie them to 0). For $f_2=\Pi M(0,1,2,5)$ the maxterm indices are the zeros, so the ones are m3, m4, m6, m7: $I_3=I_4=I_6=I_7=1$, $I_0=I_1=I_2=I_5=0$.

   8:1 MUX data inputs (select $S_2S_1S_0=ABC$; X = don’t-care)

   Function	$I_0$	$I_1$	$I_2$	$I_3$	$I_4$	$I_5$	$I_6$	$I_7$
   $f_1$	0	1	0	X (0)	1	0	1	X (0)
   $f_2$	0	0	0	1	1	0	1	1

   $$\boxed{f_1:\;(I_0,\dots,I_7)=(0,1,0,X,1,0,1,X),\qquad f_2:\;(I_0,\dots,I_7)=(0,0,0,1,1,0,1,1)}$$ For reference, the minimal gate forms are $f_2=A\overline{C}+BC$ and, taking don’t-care m3 as 1 so that m1 and m3 merge, $f_1=A\overline{C}+\overline{A}C=A\oplus C$ (without the don’t-cares it would be $A\overline{C}+\overline{A}\,\overline{B}C$).
2. Part (b), 4:1 MUX. Take the select lines $S_1S_0 = A\,B$ and reduce each function to a residue in $C$ per select address:

   MUX data-input residues (select = $AB$)

   $AB$	$I$	$f_1$ data	$f_2$ data
   00	$I_0$	$C$	$0$
   01	$I_1$	$0$	$C$
   10	$I_2$	$\overline{C}$	$\overline{C}$
   11	$I_3$	$\overline{C}$	$1$

   Each MUX needs only $C$ and $\overline{C}$ (one inverter) at its data inputs. For $f_1$ the don’t-cares sit at $AB=01$ (m3) and $AB=11$ (m7); the residues above take both as 0, which is valid. Using them instead (m3 = 1, m7 = 0) gives $I_1=C$ and the data word $(C,\,C,\,\overline{C},\,\overline{C})$, i.e. $f_1=A\oplus C$ with $B$ unused — an equally correct and more regular alternative.
3. Part (c), 3:8 decoder + OR. Drive the decoder address with $A_2A_1A_0=A\,B\,C$. Expand each function to minterms: $f_1=A(\overline{B}+C)=\Sigma m(4,5,7)$; $f_2=\Sigma m(0,3,4)$; $f_3=ABC+B\overline{C}=\Sigma m(2,6,7)$. Each output is the OR of its active-high decoder lines: $$\boxed{f_1=Y_4{+}Y_5{+}Y_7,\quad f_2=Y_0{+}Y_3{+}Y_4,\quad f_3=Y_2{+}Y_6{+}Y_7}$$